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Standard X

Mathematics

Pythagoras Theorem

A harbour lie...

Question

$A$ harbour lies in a direction $60^{0}$ south of west from a fort and at a distance $30 k m$ from it, a ship sets out from the harbour at noon and sails due east at $10 k m$ an hour. The ship will be $70 k m$ from the fort at?

7 p . m

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8 p . m

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5 p . m

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10 p . m

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Solution

The correct option is B $8 p . m$
we can see that , $a = 30 c o s 60^{o} = 15, h = 30 s i n 60^{o} = 15 \sqrt{3}$

Now, using Pythagoras Theorem,
$h^{2} + b^{2} = 4900 \Rightarrow b = \sqrt{4900 - (15 \sqrt{3})^{2}} = \sqrt{4900 - 675} = \sqrt{4225} = 65$

Now, $a + b = 65 + 15 = 80$
So, it will take $t = \frac{80}{10} = 8 H r s$

and $N o o n = 12.00 p m$

So, After 8 Hrs , Time will be $8.00 p m$

therefore Answer is $B$

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A harbour lies in a direction 600 south of west from a fort and at a distance 30km from it, a ship sets out from the harbour at noon and sails due east at 10km an hour. The ship will be 70km from the fort at?

The correct option is B 8 p.mwe can see that , a=30cos60o=15,h=30sin60o=15√3Now, using Pythagoras Theorem, h2+b2=4900⇒b=√4900−(15√3)2=√4900−675=√4225=65 Now, a+b=65+15=80So, it will take t=8010=8Hrsand Noon=12.00pmSo, After 8 Hrs , Time will be 8.00pmtherefore Answer is B

$A$ harbour lies in a direction $60^{0}$ south of west from a fort and at a distance $30 k m$ from it, a ship sets out from the harbour at noon and sails due east at $10 k m$ an hour. The ship will be $70 k m$ from the fort at?